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What will be the maximum sum of 44, 42, 40,…. ?
508
500
506
504
502
508
500
506
504
502
Solution
In order to maximize sum of 44, 42, 40,…. we consider till 2 because the next numbers will be negative and they will decrease the sum.
Hence, 44 + 42 + 40 +…….+ 2
They are in Arithmetic progression with common difference = 2
Number of terms = 44/2 =22
Hence, 44 + 42 + 40 +…….+ 2
= 11(4 + 42)
= 11 × 46 = 506
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Keynote
For an Arithmetic series: a, (a + d), (a + 2d) , (a + 3d),….
Common difference = d
First term = a
nth term = an = a + (n − 1)d
Sum of the arithmetic series:
a + (a + d) + (a + 2d) + (a + 3d),…….+ an
Study the given information carefully and answer the following questions Twelve people are sitting in two parallel rows containing six people each, in such a way that there is an equal distance between adjacent persons. In row-1 P, Q, R, S, T and V are seated and all of them are facing South. In row-2 A, B, C, D, E and F are seated and all of them are facing…
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